Electric Circuits

Review: Ohm's Law and Circuit Rules

V = IR P = VI = I²R = V²/R

In a series circuit: same current everywhere; resistances add (R_total = R₁ + R₂ + …); voltages add up to the supply voltage.
In a parallel circuit: same voltage across each branch; currents add; 1/R_total = 1/R₁ + 1/R₂ + …

EMF and Internal Resistance

A real battery has an electromotive force (EMF) ε — the work done per unit charge by the battery — and an internal resistance r that opposes current inside the battery itself.

ε = V_terminal + Ir
V_terminal = ε − Ir
Lost volts = Ir

Symbol	Quantity	Unit
ε	EMF (electromotive force)	Volt (V)
V_terminal	Terminal potential difference	Volt (V)
I	Current through battery	Ampere (A)
r	Internal resistance	Ohm (Ω)

Open circuit (I = 0): V_terminal = ε — a voltmeter across the battery reads the EMF.
Short circuit (R_external = 0): V_terminal = 0; I_max = ε/r — dangerous!
As more current is drawn, V_terminal drops. The battery appears to "weaken" under load.

Graph method: If you vary the external resistance R and plot V_terminal (y-axis) vs I (x-axis), you get a straight line: V = ε − Ir. The y-intercept gives ε; the gradient (slope) gives −r.

Kirchhoff's Laws

Kirchhoff's Current Law (KCL): At any junction (node) in a circuit, the sum of currents flowing in equals the sum of currents flowing out.

ΣI_in = ΣI_out

Kirchhoff's Voltage Law (KVL): The algebraic sum of all potential differences around any closed loop in a circuit equals zero.

ΣV = 0 (around any closed loop)

Worked example: Battery ε = 12 V, r = 1 Ω. External: R₁ = 3 Ω in series with the parallel combination of R₂ = 6 Ω and R₃ = 12 Ω.
Step 1: R_parallel = (6×12)/(6+12) = 4 Ω
Step 2: R_external = R₁ + R_parallel = 3 + 4 = 7 Ω
Step 3: I_total = ε/(R_external + r) = 12/(7+1) = 1.5 A
Step 4: V_terminal = ε − Ir = 12 − (1.5)(1) = 10.5 V
Step 5: V_R1 = IR₁ = 1.5 × 3 = 4.5 V; V_parallel = 10.5 − 4.5 = 6 V
Step 6: I_R2 = 6/6 = 1 A; I_R3 = 6/12 = 0.5 A ✓ (sum = 1.5 A)

Power and Energy

P = VI = I²R = V²/R (Watts)
E = Pt = VIt (Joules)
1 kWh = 3.6 × 10⁶ J

The resistor with the highest resistance in a series circuit dissipates the most power (P = I²R, same I). In a parallel circuit, the resistor with the lowest resistance dissipates the most power (P = V²/R, same V).

IEB Extension — Multi-loop Circuits & Special Theorems

Multi-loop circuits: Set up KVL equations for each independent loop. Choose a current direction for each loop (convention: clockwise). For each loop: Σ(EMF) = Σ(IR). Solve simultaneous equations.

Wheatstone Bridge: A circuit of four resistors in a diamond arrangement with a galvanometer across the middle. The bridge is balanced (galvanometer reads zero) when: R₁/R₂ = R₃/R₄. Used to measure unknown resistances precisely.

Maximum power transfer theorem: Maximum power is transferred from a source (with internal resistance r) to a load when the load resistance R_load = r (internal resistance). At this point, efficiency is 50%.

Show all working. Include units in every numerical answer.

Question 1

A student varies the external resistance connected to a battery and records the terminal voltage and current. The results give a straight-line graph with y-intercept 9.0 V and x-intercept 4.5 A. (a) State the EMF of the battery. (b) Calculate the internal resistance. (c) Calculate the terminal voltage when the current is 2.0 A.

Question 2

A battery (ε = 15 V, r = 0.5 Ω) is connected to an external circuit consisting of R₁ = 2 Ω in series with a parallel combination of R₂ = 4 Ω and R₃ = 12 Ω. Calculate: (a) the equivalent external resistance, (b) the total current from the battery, (c) the terminal voltage, (d) the current through R₂ and R₃, and (e) the voltage across R₁.

Question 3

In the circuit from Question 2, calculate the power dissipated in each of R₁, R₂, and R₃. Which resistor dissipates the most power? Explain why, using the relevant power formula.

Question 4

A 2 kW electric heater runs for 3 hours per day for 30 days. (a) Calculate the energy consumed in joules. (b) If electricity costs R2.50 per kWh, calculate the total cost.

Question 5

A battery of unknown EMF and internal resistance is connected to a 10 Ω resistor. The terminal voltage is 8.0 V and the current is 0.8 A. (a) Calculate the internal resistance. (b) When connected to a 5 Ω resistor instead, predict the new current and terminal voltage. (c) Explain why the terminal voltage is lower when a smaller resistance is used.

Question 6 — Finding EMF and r from Measured Data

A learner varies the external resistor connected to a battery and measures the current drawn and the resulting terminal voltage each time:

Current I (A)	0.5	1.0	1.5	2.0
Terminal voltage V (V)	8.5	7.0	5.5	4.0

(a) Calculate the change in V for each 0.5 A increase in I. What does this confirm about the relationship between V and I?
(b) Using any two rows of the table, calculate the internal resistance r of the battery (the magnitude of the gradient of V vs I).
(c) Using your value of r and one row of data, calculate the EMF of the battery (the value V would have at I = 0).
(d) Using the pattern in the table (not just the EMF/r equation), predict the terminal voltage when I = 3.0 A.

Review: Ohm's Law and Circuit Rules

EMF and Internal Resistance

Kirchhoff's Laws

Power and Energy

Internal Resistance — Battery Model with Load